Difference between revisions of "Compactness in Metric Spaces"

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<div class="image-container aligncenter"><img src="http://mathonline.wdfiles.com/local--files/boundedness-of-compact-sets-in-a-metric-space/Screen%20Shot%202015-10-05%20at%209.56.39%20PM.png" alt="Screen%20Shot%202015-10-05%20at%209.56.39%20PM.png" class="image" /></div>
 
<div class="image-container aligncenter"><img src="http://mathonline.wdfiles.com/local--files/boundedness-of-compact-sets-in-a-metric-space/Screen%20Shot%202015-10-05%20at%209.56.39%20PM.png" alt="Screen%20Shot%202015-10-05%20at%209.56.39%20PM.png" class="image" /></div>
 
<ul>
 
<ul>
<li><strong>Proof:</strong> For a fixed <span class="math-inline"><math>x_0 \in S</math></span> and for <span class="math-inline"><math>r &gt; 0</math></span>, consider the ball centered at <span class="math-inline"><math>x_0</math></span> with radius <span class="math-inline"><math>r</math></span>, i.e., <span class="math-inline"><math>B(x_0, r)</math></span>. Let <span class="math-inline"><math>\mathcal F</math></span> denote the collection of balls centered at <span class="math-inline"><math>x_0</math></span> with varying radii <span class="math-inline"><math>r &gt; 0</math></span>:</li>
+
<li><strong>Proof:</strong> For a fixed <span class="math-inline"><math>x_0 \in S</math></span> and for <span class="math-inline"><math>r > 0</math></span>, consider the ball centered at <span class="math-inline"><math>x_0</math></span> with radius <span class="math-inline"><math>r</math></span>, i.e., <span class="math-inline"><math>B(x_0, r)</math></span>. Let <span class="math-inline"><math>\mathcal F</math></span> denote the collection of balls centered at <span class="math-inline"><math>x_0</math></span> with varying radii <span class="math-inline"><math>r > 0</math></span>:</li>
 
</ul>
 
</ul>
<div style="text-align: center;"><math>\begin{align} \quad \mathcal F = \{ B(x_0, r) : r &gt; 0 \} \end{align}</math></div>
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<div style="text-align: center;"><math>\begin{align} \quad \mathcal F = \{ B(x_0, r) : r > 0 \} \end{align}</math></div>
 
<ul>
 
<ul>
<li>It should not be hard to see that <span class="math-inline"><math>\mathcal F</math></span> is an open covering of <span class="math-inline"><math>S</math></span>, since for all <span class="math-inline"><math>s \in S</math></span> we have that <span class="math-inline"><math>d(x_0, s) = r_s &gt; 0</math></span>, so <span class="math-inline"><math>s \in B(x_0, r_s) \in \mathcal F</math></span>.</li>
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<li>It should not be hard to see that <span class="math-inline"><math>\mathcal F</math></span> is an open covering of <span class="math-inline"><math>S</math></span>, since for all <span class="math-inline"><math>s \in S</math></span> we have that <span class="math-inline"><math>d(x_0, s) = r_s > 0</math></span>, so <span class="math-inline"><math>s \in B(x_0, r_s) \in \mathcal F</math></span>.</li>
 
</ul>
 
</ul>
 
<ul>
 
<ul>
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<div style="text-align: center;"><math>\begin{align} \quad S \subseteq \bigcup_{k=1}^{p} B(x_0, r_k) \end{align}</math></div>
 
<div style="text-align: center;"><math>\begin{align} \quad S \subseteq \bigcup_{k=1}^{p} B(x_0, r_k) \end{align}</math></div>
 
<ul>
 
<ul>
<li>Each of the open balls in the open subcovering <span class="math-inline"><math>\mathcal F^*</math></span> is centered at <span class="math-inline"><math>x_0</math></span> with <span class="math-inline"><math>r_1, r_2, ..., r_p &gt; 0</math></span>. Since the set <span class="math-inline"><math>\{ r_1, r_2, ..., r_p \}</math></span> is a finite set, there exists a maximum value. Let:</li>
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<li>Each of the open balls in the open subcovering <span class="math-inline"><math>\mathcal F^*</math></span> is centered at <span class="math-inline"><math>x_0</math></span> with <span class="math-inline"><math>r_1, r_2, ..., r_p > 0</math></span>. Since the set <span class="math-inline"><math>\{ r_1, r_2, ..., r_p \}</math></span> is a finite set, there exists a maximum value. Let:</li>
 
</ul>
 
</ul>
 
<div style="text-align: center;"><math>\begin{align} \quad r_{\mathrm{max}} = \max \{ r_1, r_2, ..., r_p \} \end{align}</math></div>
 
<div style="text-align: center;"><math>\begin{align} \quad r_{\mathrm{max}} = \max \{ r_1, r_2, ..., r_p \} \end{align}</math></div>
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<li>Hence <span class="math-inline"><math>S</math></span> is bounded. <span class="math-inline"><math>\blacksquare</math></span></li>
 
<li>Hence <span class="math-inline"><math>S</math></span> is bounded. <span class="math-inline"><math>\blacksquare</math></span></li>
 
</ul>
 
</ul>
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==Licensing==
 
==Licensing==
 
Content obtained and/or adapted from:
 
Content obtained and/or adapted from:
 
* [http://mathonline.wikidot.com/compact-sets-in-a-metric-space Compact Sets in a Metric Space, mathonline.wikidot.com] under a CC BY-SA license
 
* [http://mathonline.wikidot.com/compact-sets-in-a-metric-space Compact Sets in a Metric Space, mathonline.wikidot.com] under a CC BY-SA license

Revision as of 15:55, 8 November 2021

Compact Sets in a Metric Space

If is a metric space and then a cover or covering of is a collection of subsets in such that:

Furthermore, we said that an open cover (or open covering) is simply a cover that contains only open sets.

We also said that a subset is a subcover/subcovering (or open subcover/subcovering if is an open covering) if is also a cover of , that is:

We can now define the concept of a compact set using the definitions above.

Definition: Let be a metric space. The subset is said to be Compact if every open covering of has a finite subcovering of .

In general, it may be more difficult to show that a subset of a metric space is compact than to show a subset of a metric space is not compact. So, let's look at an example of a subset of a metric space that is not compact.

Consider the metric space where is the Euclidean metric and consider the set . We claim that this set is not compact. To show that is not compact, we need to find an open covering of that does not have a finite subcovering. Consider the following open covering:

Clearly is an infinite subcovering of and furthermore:

Let be a finite subset of containing elements. Then:

Let . Then due to the nesting of the open covering , we see that:

But for we need . But , so and , so . Therefore any finite subset of cannot cover . Hence, is not compact.

Boundedness of Compact Sets in a Metric Space

Recall that if is a metric space then a subset is said to be compact in if for every open covering of there exists a finite subcovering of .

We will now look at a rather important theorem which will tell us that if is a compact subset of then we can further deduce that is also a bounded subset.

Theorem 1: If be a metric space and is a compact subset of then is bounded.

  • Proof: For a fixed and for , consider the ball centered at with radius , i.e., . Let denote the collection of balls centered at with varying radii :
  • It should not be hard to see that is an open covering of , since for all we have that , so .
  • Now since is compact and since is an open covering of , there exists a finite open subcovering subset that covers . Since is finite, we have that:
  • And by definition covers so:
  • Each of the open balls in the open subcovering is centered at with . Since the set is a finite set, there exists a maximum value. Let:
  • Then for all we have that and therefore:
  • Hence is bounded.


Licensing

Content obtained and/or adapted from: